A manufacturer produces two models of bikes-model X and model Y. Model X takes a 6 man-hours to make per unit, while model Y takes 10 man hours per unit. There is a total of 450 man-hour available per week. Handling and marketing costs are Rs.2000 and Rs.1000 per unit for models X and Y, respectively. The total funds available for these purposes are Rs.80000 per week. Profits per unit for models X and Y are Rs.1000 and Rs.500, respectively. How many bikes of each model should the manufacturer produce, so as to yield a maximum profit? Find the maximum profit.

Let the manufacturer produces $x$ number of models X and $y$ number of model Y bikes.

Model X takes a 6 man-hours to make per unit and model Y takes a 10 man-hours to make per unit.

There is total of 450 man-hour available per week.

$\therefore$ $6x + 10y \le 450$ …..(i)

$\Rightarrow$ $3x + 5y \le 225$

For models X and Y, handling and marketing costs are Rs.2000 and Rs.1000,

respectively, total funds available for these purposes are Rs.80000 per week.

$\therefore$ $2000x + 1000y \le 80000$
$\Rightarrow$ $2x + y \le 80$ …..(ii)

Also, $x \ge 0,y \ge 0$

Hence, the profits per unit for models X and Y are Rs.1000 and Rs.500, respectively.

$\therefore$ Required LPP is

Maximise $Z = 1000x + 500y$

Subject to, $3x + 5y \le 225,$ $2x + y \le 80,$ $x \ge 0,$ $y \ge 0$

From the shaded feasible region, it is clear

that coordinates of corner points are (0,0) , (40,0),(25,30) and (0,45).

On solving $3x + 5y = 225$ and $2x + y = 80$, we get
$x = 25,y = 30$

So, the manufacturer should produce 25 bikes of model X and 30 bikes of model Y to get A maximum profit of Rs.40000.